2003 OIM Problems/Problem 1

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Problem

a) There are two sequences, each of 2003 consecutive integers, and a board with 2 rows and 2003 columns

Decide if it is always possible to distribute the numbers of the first sequence in the first row and those of the second sequence in the second row, in such a way that the results obtained by adding the two numbers in each column form a new sequence of 2003 consecutive numbers.

b) What if 2003 is replaced by 2004?

In both a) and b), if the answer is affirmative, explain how you would distribute the numbers, and if the answer is negative, justify why.

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

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We have,

Sequence 1 (First Row): a,a+1,a+2,,a+2002

Sequence 2 (Second Row): b,b+1,b+2,,b+2002

We need to arrange these numbers such that the sum in each column forms a sequence of 2003 consecutive numbers.

Let Si be the sum of the numbers in column i, where i ranges from 0 to 2002.

\[S_i = \text{First Row}_i + \text{Second Row}_i\]

Suppose we directly distribute the numbers into each row:

First Row: a+i

Second Row: b+i

Then the sum in each column is:

\[S_i = (a + i) + (b + i) = a + b + 2i\]

We want Si to form a new sequence of 2003 consecutive numbers.

Suppose Si forms a sequence starting at x:

\[S_i = x + i\]

For Si=a+b+2i, we need to find appropriate a and b such that a+b+2i=x+i for all i.

This implies:

\[a + b + i = x\]

This is impossible for all i because a and b are fixed, while i is variable.

Therefore, we cannot distribute the numbers in such a way that the sums in each column form a new sequence of 2003 consecutive numbers.

The same analysis applies to 2004 integers:

Sequence 1 (First Row): a,a+1,a+2,,a+2003

Sequence 2 (Second Row): b,b+1,b+2,,b+2003

Following similar steps, the sum in each column Si would be:

\[S_i = (a + i) + (b + i) = a + b + 2i\]

Again, it is impossible to form a new sequence of 2004 consecutive numbers.

For both 2003 and 2004 consecutive integers, it is impossible to distribute the numbers in the sequences into two rows in such a way that the sums of the numbers in each column form a new sequence of consecutive numbers.

Thus, the answer is:

For both 2003 and 2004 integers, it is not possible.

\[\boxed{\text{Not possible}}\]

See also